(unesp)um triângulo tem vertices P=(2,1),q=(2,5)e R=(x,4). sabendo-se que a área do triângulo é 20,calcule a abscissa do ponto R.
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Boa tarde Caroline!
Solução!
Vamos montar um determinante para calcularmos o valo de x que tem essa forma.
![Area= \dfrac{1}{2}\times|D|,sendo~~D= \begin{vmatrix}
x_{1} & y_{1}&1 \\
x_{2} & y_{2}&1 \\
x_{3} &y _{3} &1
\end{vmatrix}\\\\\\\\\\
P(2,1)\\\\\
Q(2,5)\\\\\\
R(x,4)\\\\\\
Area=20](https://tex.z-dn.net/?f=Area%3D+%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%7CD%7C%2Csendo%7E%7ED%3D+%5Cbegin%7Bvmatrix%7D+%0A+++x_%7B1%7D++%26amp%3B+y_%7B1%7D%26amp%3B1+%5C%5C+%0A+++x_%7B2%7D+%26amp%3B+y_%7B2%7D%26amp%3B1++%5C%5C%0A+x_%7B3%7D+%26amp%3By+_%7B3%7D+%26amp%3B1%0A++%5Cend%7Bvmatrix%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AP%282%2C1%29%5C%5C%5C%5C%5C%0AQ%282%2C5%29%5C%5C%5C%5C%5C%5C%0AR%28x%2C4%29%5C%5C%5C%5C%5C%5C%0AArea%3D20 "Area= \dfrac{1}{2}\times|D|,sendo~~D= \begin{vmatrix}
x_{1} & y_{1}&1 \\
x_{2} & y_{2}&1 \\
x_{3} &y _{3} &1
\end{vmatrix}\\\\\\\\\\
P(2,1)\\\\\
Q(2,5)\\\\\\
R(x,4)\\\\\\
Area=20")
![\begin{vmatrix}
2 & 1&1&2&1 \\
2 & 5&1&2&5 \\
x &4 &1&x&4
\end{vmatrix}=20\\\\\\\\\\
Multiplicando~~as~~diagonais~~principais~~e~~as~~secundarias.\\\\\\\\\
(10+x+8)+(-5x-8-2)=20\\\\\\
x-5x+10+8-8-2=20\\\\\
-4x+8=20\\\\\\
-4x=20-8\\\\\
-4x=12\\\\\
x= \dfrac{12}{-4}\\\\\\
x=-3](https://tex.z-dn.net/?f=%5Cbegin%7Bvmatrix%7D+%0A+++2++%26amp%3B+1%26amp%3B1%26amp%3B2%26amp%3B1+%5C%5C+%0A+++2+%26amp%3B+5%26amp%3B1%26amp%3B2%26amp%3B5++%5C%5C%0A+x+%26amp%3B4+%26amp%3B1%26amp%3Bx%26amp%3B4%0A++%5Cend%7Bvmatrix%7D%3D20%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AMultiplicando%7E%7Eas%7E%7Ediagonais%7E%7Eprincipais%7E%7Ee%7E%7Eas%7E%7Esecundarias.%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%2810%2Bx%2B8%29%2B%28-5x-8-2%29%3D20%5C%5C%5C%5C%5C%5C%0Ax-5x%2B10%2B8-8-2%3D20%5C%5C%5C%5C%5C%0A-4x%2B8%3D20%5C%5C%5C%5C%5C%5C%0A-4x%3D20-8%5C%5C%5C%5C%5C%0A-4x%3D12%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B12%7D%7B-4%7D%5C%5C%5C%5C%5C%5C%0Ax%3D-3+%0A "\begin{vmatrix}
2 & 1&1&2&1 \\
2 & 5&1&2&5 \\
x &4 &1&x&4
\end{vmatrix}=20\\\\\\\\\\
Multiplicando~~as~~diagonais~~principais~~e~~as~~secundarias.\\\\\\\\\
(10+x+8)+(-5x-8-2)=20\\\\\\
x-5x+10+8-8-2=20\\\\\
-4x+8=20\\\\\\
-4x=20-8\\\\\
-4x=12\\\\\
x= \dfrac{12}{-4}\\\\\\
x=-3")
Boa tarde!
Bons estudos!
Solução!
Vamos montar um determinante para calcularmos o valo de x que tem essa forma.
Boa tarde!
Bons estudos!
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